Nesbitt's Inequality: Alternative Proof

I recently came across the well-known Nesbitt's inequality while reading about famous mathematical relations. If you visit the Wikipedia page for this inequality, there are six different proofs for it. I then proceeded to prove it in a different way, that is, the method I used was not articulated on the Wikipedia page. The proof I wrote is below, I thought Brilliant users would find it somewhat interesting (if you find any errors in it, please point them out as I'm trying to improve my proof writing skills):

Nesbitt's Inequality:

\(\forall\ a,b,c \in \mathbb{R}^{+} : \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{3}{2}\)

Proof:

Let \(f(a,b,c)= \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}\). We will show that \(f(a,b,c)\) has a minimum when \(a=b=c\) and that all other extrema imply that at least one of \(a,b,c \notin \mathbb{R}^{+}\).

All maxima and minima occur when \(\nabla f(a,b,c)=\vec{0}\). Hence we find the partial derivatives of \(f(a,b,c)\):

\(\frac{\partial f}{\partial a} = \frac{1}{b+c} -\frac{b}{(a+c)^2}-\frac{c}{(a+b)^2}\)

\(\frac{\partial f}{\partial b}= -\frac{a}{(b+c)^2}+\frac{1}{a+c}-\frac{c}{(a+b)^2}\)

\(\frac{\partial f}{\partial c}= -\frac{a}{(b+c)^2}-\frac{b}{(a+c)^2}+\frac{1}{a+b}\)

We now set \(\nabla f(a,b,c)=\vec{0}\):

\(0=\frac{1}{b+c} -\frac{b}{(a+c)^2}-\frac{c}{(a+b)^2}\)

\(0=-\frac{a}{(b+c)^2}+\frac{1}{a+c}-\frac{c}{(a+b)^2}\)

\(0=-\frac{a}{(b+c)^2}-\frac{b}{(a+c)^2}+\frac{1}{a+b}\)

Solving this non-linear system:

\(\frac{1}{a+c}+\frac{b}{(a+c)^2} = \frac{1}{a+b} + \frac{c}{(a+b)^2}\) \(\Rightarrow\) \((a+c)^2=(a+b)^2\) \(\Rightarrow\) \(c=-a \pm (a+b)\) \(\Rightarrow\) \(c=b\) or \(c=-2a-b\)

\(c=-2a-b\) implies \(c \notin \mathbb{R}^{+}\), since we assume \(a,b \in \mathbb{R}^{+}\). Hence we choose \(c=b\).

\(\frac{1}{b+c}-\frac{b}{(a+c)^2}=\frac{1}{a+c}-\frac{a}{(b+c)^2}\) \(\Rightarrow\) \(\frac{a+b+c}{(b+c)^2}=\frac{a+b+c}{(a+c)^2}\) \(\Rightarrow\) \(a=-b \pm 2b\) \(\Rightarrow\) \(a=-3b\) or \(a=b\).

\(a=-3b\) implies \(a \notin \mathbb{R}^{+}\) since we assume \(b \in \mathbb{R}^{+}\). Then \(a=b\) \(\Rightarrow\) \(a=b=c\).

Since all other solutions of this system require that at least one of \(a,b,c\) is negative, the only solution such that \(a,b,c \in \mathbb{R}^{+}\) is \(a=b=c\), which actually represents infinitely many solutions. Note that \(f(a,b,c)=\frac{3}{2}\) if \(a=b=c\). Also note that this must be a minimum, because if we let \(a=1,b=2,c=3\), then \(f(a,b,c)= 1.7 \geq \frac{3}{2}\). Then it follows that \(\forall\ a,b,c \in \mathbb{R}^{+} : \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{3}{2}\), which was to be proved.

QED

Note by Ethan Robinett
3 years, 8 months ago

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I was actually the one who wrote the sixth proof on wikipedia :P

Daniel Liu - 3 years, 8 months ago

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Lol that's actually pretty cool

Ethan Robinett - 3 years, 8 months ago

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There can also be a seventh proof using convex functions but I am too lazy to write it. : |

Arif Ahmed - 3 years, 8 months ago

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